Expression for Angular Frequency

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  • #1
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How is this expression for angular frequency in an L.C.R. circuit derived-

w = (1/LC - R2/4L2)1/2

(I came across this while solving a problem)
 
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  • #2
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You solve the differential equation for kirchoff's loop rule.
 
  • #3
tiny-tim
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Hi Abdul! :smile:
How does this expression for angular frequency in an L.C.R. circuit derived-

w = (1/LC - R2/4L2)1/2

(I came across this while solving a problem)

Well, without knowing anything about circuits, we can guess that it probably has something to do with the fact that that obviously looks like a determinant,

of Lx2 + Rx + 1/C = 0 …​

can you get a differential equation that looks like that, maybe for current? :wink:
 
  • #4
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I got the same equation for w in place of x. On solving,

w = (-R +/- (R2 -4L/C)1/2)/2L

I did not get any differential equation.
 
  • #5
tiny-tim
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I got the same equation for w in place of x. On solving,

w = (-R +/- (R2 -4L/C)1/2)/2L

Are you sure? that's not the same ω as in your original post …
How is this expression for angular frequency in an L.C.R. circuit derived-

w = (1/LC - R2/4L2)1/2

I suspect that the R/2L is a separate exponential part

(my differential equation in I was from combining CVC = Q (so CdVC/dt = I), VR = IR, VL = LdI/dt, and d/dt(VC + VR + VL) = 0)
 
  • #6
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(my differential equation in I was from combining CVC = Q (so CdVC/dt = I), VR = IR, VL = LdI/dt, and d/dt(VC + VR + VL) = 0)

That will give a second order differential equation, right?
Your expression does not contain any w.
 
  • #7
tiny-tim
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It gives I as a function of t …

part of that solution will be harmonic, and ω will be the frequency of the harmonic part. :smile:
 
  • #8
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I got
I/C + RdI/dt + Ld2I/dt2 = 0

How to solve this?
 
  • #9
tiny-tim
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Hi Abdul! :smile:

(just got up :zzz: …)
I got
I/C + RdI/dt + Ld2I/dt2 = 0

How to solve this?

Don't you know the standard method for these equations? …

write D = d/dt, so the equation becomes

(LD2 + RD + 1/C)I = 0,

which you can factor to

(D - R/2L + i√(1/LC + R2/4L2))(D - R/2L - i√(1/LC + R2/4L2))I = 0 …

carry on from there :smile:
 
  • #10
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Its d2I/dt2 not (dI/dt)2 :wink:
 
  • #11
tiny-tim
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That's ok, that is what D2 means …

D2(I) = D(D(I)) = d/dt(d/dt(I)) = d/dt(dI/dt) = d2I/dt2 :smile:
 
  • #12
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:surprised

I have solved questions in D.E. like -
1 + (dy/dx)2 = xdy/dx

by taking dy/dx = p

Do we use p2 for both (dy/dx)2 and d2y/dx2 ?
 
  • #13
tiny-tim
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no …

that p equation has an xdy/dx,

the D is for equations with just dy/dx :smile:
 

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